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Lie's second theorem says that if $G$ is a connected simply connected Lie group with Lie algebra $\mathfrak g$, then the functor of "differentiation" from the category $\mathrm{Rep}^f(G)$ of finite-dimensional representations of $G$ to the category $\mathrm{Rep}^f(\mathfrak g)$ of finite-dimensional representations of $\mathfrak g$ is an equivalence of categories. The inverse functor deserves to be called "integration".

There are actually two versions of this theorem, one for real (necessarily analytic) groups and real representations, and one for complex (holomorphic) groups and complex representations. In both cases the proof is essentially no worse than the existence of solutions to ODEs.

My question is about "Lie monoids", which are monoid objects in (real or complex) manifolds that are not necessarily groups. Any Lie monoid $M$ has a maximal sub-Lie group $M^\times$ (its group of invertible elements), and the infinitesimal neighborhood of $e\in M$ certainly cannot see outside $M^\times$. Thus one can define $\mathrm{Lie}(M) = \mathrm{Lie}(M^\times)$. Note that, unlike in the group case, this is no longer the same as the Lie algebra of left-translation-invariant vector fields. I do still have a "differentiation" functor $\mathrm{Rep}^f(M) \to \mathrm{Rep}^f(\mathrm{Lie}(M))$.

Are there conditions on $M$, short of demanding that it be a group, in which I might still have an equivalence of categories?

What if instead I used the Lie algebra of left-translation-invariant vector fields?


Addendum: The comments have brought up a lot of interesting mathematics, but also suggest that perhaps I didn't make my question clear. Here is a version of what I meant to ask:

Among the monoids $M$ with a given Lie algebra $\mathfrak g$, some of them have the property that $\mathrm{Rep}^f(M) \to \mathrm{Rep}^f(\mathfrak g)$ is an equivalence. (Provided $M$ is connected, I think it suffices for every $\mathfrak g$-module to extend to an $M$-module.) This class contains all the connected simply-connected groups with Lie algebra $\mathfrak g$ (of which there is, of course, only one up to unique isomorphism). It is not limited just to the connected simply-connected groups: if $\mathfrak g = \mathfrak{sl}(2,\mathbb R)$, then this class of monoids contains $\mathrm{SL}(2,\mathbb R)$, for example. Might it also contain non-group monoids?

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    $\begingroup$ Consider the two embeddings $GL(n) \hookrightarrow M_n$, $GL(n) \stackrel{-1}\to GL(n) \hookrightarrow M_n$. What do you want such a criterion to say, in these examples? Is the Lie algebra side perhaps about pairs $(\mathfrak g,C)$ where $C$ is a subcone of $\mathfrak g$'s positive Weyl chamber? $\endgroup$ – Allen Knutson May 2 '15 at 1:34
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    $\begingroup$ Did you check the quite well developed theory of Lie subsemigroups of Lie groups, whose infinitesimal objects are cones in Lie algebras. See for example the book: MR1032761 Reviewed Hilgert, Joachim; Hofmann, Karl Heinrich; Lawson, Jimmie D. Lie groups, convex cones, and semigroups. Oxford Mathematical Monographs. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1989. xxxviii+645 pp. ISBN: 0-19-853569-4 (Reviewer: K. Strambach) $\endgroup$ – Peter Michor May 2 '15 at 9:34
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    $\begingroup$ It might help us to give you references if you were to either give the actual definition of a Lie monoid or give a reference in which 'Lie monoid' is defined. The description you gave isn't precise enough for us to decide what a 'Lie monoid' is supposed to be. I would have thought that you want every Lie subsemigroup of a Lie group to be a Lie monoid, but when you write that they are 'monoid objects in (real or complex) manifolds', that makes it seem that you want them to be manifolds, which Lie subsemigroups generally are not (unless you want to count manifolds with corners as manifolds). $\endgroup$ – Robert Bryant May 3 '15 at 9:11
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    $\begingroup$ @TheoJohnson-Freyd: OK, then I'm a bit mystified as to how Peter's references might be helpful in understanding Lie monoids. Allen's example of the Lie monoids $M_n(\mathbb{F})$ (where $\mathbb{F}$ could be any of $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$) shows that you can't use the Lie algebra of the Lie group of the invertible elements to tell you much about the noninvertible elements (if any) in the Lie monoid. I suppose you could ask for the classification of all the connected Lie monoids that contain a given connected Lie group as its (open) submonoid of invertible elements. $\endgroup$ – Robert Bryant May 4 '15 at 0:52
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    $\begingroup$ In your "standard example", by "all" Lie algebra representations you mean (continuous and) finite-dimensional, apparently? $\endgroup$ – Allen Knutson May 5 '15 at 5:22

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